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Difference between revisions of "Mock AIME 4 2006-2007 Problems/Problem 7"
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==Problem==
 
==Problem==
 
Find the remainder when <math>3^{3^{3^3}}</math> is divided by 1000.
 
Find the remainder when <math>3^{3^{3^3}}</math> is divided by 1000.
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==Solution==
 
==Solution==
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Using the [[Carmichael function]], we have <math>\lambda(1000)=100</math>, so <math>3^{100}=1\pmod{1000}</math>. Therefore, letting <math>N=3^{3^3}</math>, we seek to find an <math>n</math> such that <math>N\equiv n\pmod{100}</math> so that <math>3^N\equiv 3^n\pmod{1000}</math>.
  
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Fermat's little theorem comes in handy here:
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Using the Carmichael function again, we have <math>\lambda(100)=20</math>, so <math>N=3^{27}\equiv 3^7\pmod{100}\equiv 87\pmod{100}</math>. Therefore <math>n=87</math>, and so we have the following:
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<cmath>
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<math>3^{400}\equiv 1 \pmod {1000}</math>
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3^{3^{3^3}}\equiv 3^{87}\pmod{1000}.
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</cmath>
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So we are looking for <math>3^{27} \pmod {400}</math>
 
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<math>3^{27}=(3^9)^3=(27^3)^3</math>
 
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We brute force it:
 
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<math>27^3=729*27=19683</math>
 
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<math>19683 \equiv 83 \pmod {400}</math>
 
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<math>83^2=6889</math>
 
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<math>83^3 \equiv 187 \pmod {400}</math>
 
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<math>3^{27} \equiv 187 \pmod {400}</math>
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Now,
  
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<math>3^{3^{3^3}} \equiv 3^{187} \pmod {1000}</math>
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<cmath>\begin{align*}3^{87}=(3^{20})^4\cdot 3^7&\equiv 401^4\cdot 187\pmod{1000} \\
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&\equiv 601\cdot 187\pmod{1000} \\
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&\equiv \boxed{387}\pmod{1000}.
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\end{align*}</cmath>
  
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{{solution}}
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== See also==
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----
 
  
 
*[[Mock AIME 4 2006-2007 Problems/Problem 8| Next Problem]]
 
*[[Mock AIME 4 2006-2007 Problems/Problem 8| Next Problem]]
 
*[[Mock AIME 4 2006-2007 Problems/Problem 6| Previous Problem]]
 
*[[Mock AIME 4 2006-2007 Problems/Problem 6| Previous Problem]]
 
*[[Mock AIME 4 2006-2007 Problems]]
 
*[[Mock AIME 4 2006-2007 Problems]]
 +
 +
[[Category:Intermediate Number Theory Problems]]

Latest revision as of 00:15, 5 January 2010

Problem

Find the remainder when $3^{3^{3^3}}$ is divided by 1000.

Solution

Using the Carmichael function, we have $\lambda(1000)=100$, so $3^{100}=1\pmod{1000}$. Therefore, letting $N=3^{3^3}$, we seek to find an $n$ such that $N\equiv n\pmod{100}$ so that $3^N\equiv 3^n\pmod{1000}$.

Using the Carmichael function again, we have $\lambda(100)=20$, so $N=3^{27}\equiv 3^7\pmod{100}\equiv 87\pmod{100}$. Therefore $n=87$, and so we have the following: \[3^{3^{3^3}}\equiv 3^{87}\pmod{1000}.\]

Now,

\begin{align*}3^{87}=(3^{20})^4\cdot 3^7&\equiv 401^4\cdot 187\pmod{1000} \\ &\equiv 601\cdot 187\pmod{1000} \\ &\equiv \boxed{387}\pmod{1000}. \end{align*}

See also

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